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Minimum Value (Posted on 2002-06-19) Difficulty: 5 of 5
Given:
    x,y,z > 0
    xy+yz+zx <= 3/4
and
    P=x+y+z+(1/x)+(1/y)+(1/z)
Find the Minimum Value of P. Find x, y, and z when P = Minimum Value.

See The Solution Submitted by vohonam    
Rating: 3.2500 (8 votes)

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Solution possible minimum value | Comment 13 of 18 |
if x=y=z=1 then the smallest value for P is six... i can't find anyway to get P less than six even with using three different variables... if smaller than one then the fraction becomes greater, if larger than one think of it this way P=3(x+1/x) the smallest value for x+1/x is two
and to check it
also first derivative of that would be0=1-1/x^2 (P goes to zero because as a minimum value its a constant) so 1=1/x^2or x^2=1... and yeah....enough babbling from the newbie
  Posted by lyeshea on 2002-07-02 12:43:56
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